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Chapter 3: Problem 6

How many solutions will the linear system \(A \mathbf{x}=\mathbf{b}\) have if \(\mathbf{b}\) is in the column space of \(A\) and the column vectors of \(A\) are linearly dependent? Explain.

Short Answer

Expert verified
The linear system \(A \mathbf{x}=\mathbf{b}\) has infinitely many solutions when \(\mathbf{b}\) is in the column space of \(A\) and the column vectors of \(A\) are linearly dependent. This is because there is at least one solution for \(\mathbf{b}\), and by adding any scalar multiple of the non-trivial solution to the homogeneous system, we can generate an infinite number of additional solutions.

Step by step solution

01

The column space of a matrix, \(Col(A)\), is the set of all linear combinations of its column vectors. In other words, for a vector \(\mathbf{b}\) to be in the column space of a matrix \(A\), it must be possible to express \(\mathbf{b}\) as a linear combination of the columns of \(A\). Linear dependence means that one (or more) of the column vectors of a matrix can be expressed as a linear combination of the others. In other words, there is some redundancy among the column vectors of the matrix \(A\). #Step 2: Understanding the Implications of the Given Conditions#

Since \(\mathbf{b}\) is in the column space of \(A\), we can express \(\mathbf{b}\) as a linear combination of the column vectors of \(A\): \[\mathbf{b} = c_1 \mathbf{a}_1 + c_2 \mathbf{a}_2 + \cdots + c_n \mathbf{a}_n\] where \(\mathbf{a}_1, \mathbf{a}_2, \cdots, \mathbf{a}_n\) are the column vectors of \(A\) and \(c_1, c_2, \cdots, c_n\) are scalar constants. This means that there exists at least one solution to the linear system \(A \mathbf{x} = \mathbf{b}\), namely \(\mathbf{x}=\begin{bmatrix} c_1\\c_2\\ \vdots\\ c_n \end{bmatrix}\). Furthermore, since the column vectors of \(A\) are linearly dependent, there exists a non-trivial (non-zero) solution to the homogeneous system: \(A \mathbf{x} = \mathbf{0}\). #Step 3: Determine the Number of Solutions#
02

The linear system \(A \mathbf{x} = \mathbf{b}\) has at least one solution, as established in step 2. However, because the column vectors of \(A\) are linearly dependent, we can add any scalar multiple of the non-trivial solution of the homogeneous system \(A \mathbf{x} = \mathbf{0}\) to the known solution of \(A \mathbf{x} = \mathbf{b}\), and the sum will still be a solution of \(A \mathbf{x} = \mathbf{b}\): \[\mathbf{x'} = \begin{bmatrix} c_1\\c_2\\ \vdots\\ c_n \end{bmatrix} + k \mathbf{x}_{hom}\] where \(\mathbf{x}_{hom}\) is a non-trivial solution to the homogeneous system \(A \mathbf{x} = \mathbf{0}\), \(k\) is a scalar constant and \(\mathbf{x'}\) is a new solution to the system. Since we can choose any scalar for \(k\), we can produce infinitely many solutions to the given system. Thus, the linear system \(A \mathbf{x} = \mathbf{b}\) has infinitely many solutions. #Step 4: Conclusion#

The linear system \(A \mathbf{x}=\mathbf{b}\) has infinitely many solutions, given that \(\mathbf{b}\) is in the column space of the matrix \(A\) and the column vectors of \(A\) are linearly dependent. This is because there is at least one solution for \(\mathbf{b}\), and by adding any scalar multiple of the non-trivial solution to the homogeneous system, we can generate an infinite number of additional solutions.

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Most popular questions from this chapter

Let \(\mathbf{x}_{1}, \mathbf{x}_{2},\) and \(\mathbf{x}_{3}\) be linearly independent vectors in \(\mathbb{R}^{n}\) and let $$\mathbf{y}_{1}=\mathbf{x}_{1}+\mathbf{x}_{2}, \quad \mathbf{y}_{2}=\mathbf{x}_{2}+\mathbf{x}_{3}, \quad \mathbf{y}_{3}=\mathbf{x}_{3}+\mathbf{x}_{1}$$ Are \(\mathbf{y}_{1}, \mathbf{y}_{2},\) and \(\mathbf{y}_{3}\) linearly independent? Prove your answer.

Let \(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\) be linearly independent vectors in a vector space \(V .\) Show that \(\mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\) cannot \(\operatorname{span} V\)

Show that if \(U\) and \(V\) are subspaces of \(\mathbb{R}^{n}\) and \(U \cap V=\\{0\\},\) then \\[ \operatorname{dim}(U+V)=\operatorname{dim} U+\operatorname{dim} V \\]

Given \(\mathbf{x}_{1}=(1,1,1)^{T}\) and \(\mathbf{x}_{2}=(3,-1,4)^{T}:\) (a) Do \(\mathbf{x}_{1}\) and \(\mathbf{x}_{2}\) span \(\mathbb{R}^{3} ?\) Explain. (b) Let \(\mathbf{x}_{3}\) be a third vector in \(\mathbb{R}^{3}\) and \(\operatorname{set} X=\) \(\left(\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\right) .\) What condition \((\mathrm{s})\) would \(X\) have to satisfy in order for \(\mathbf{x}_{1}, \mathbf{x}_{2},\) and \(\mathbf{x}_{3}\) to form a basis for \(\mathbb{R}^{3}\) ? (c) Find a third vector \(\mathbf{x}_{3}\) that will extend the set \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}\right\\}\) to a basis for \(\mathbb{R}^{3}\).

Let \(S\) be the subspace of \(\mathbb{R}^{2}\) spanned by \(\mathbf{e}_{1}\) and let \(T\) be the subspace of \(\mathbb{R}^{2}\) spanned by \(\mathbf{e}_{2}\). Is \(S \cup T\) a subspace of \(\mathbb{R}^{2}\) ? Explain.
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